Question

In the estimation of nitrogen by Kjeldahl’s method, $2.8{\text{g}}$ for an organic compound required \[20{\text{mmoles}}\] of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ for the complete neutralization of ${\text{N}}{{\text{H}}_3}$ gas evolved. The percentage of nitrogen in the sample is:
A. $20\% $
B. $10\% $
C. $40\% $
D. $30\% $

Answer 1

Hint: The main objective of Kjeldahl’s method is quantitative determination of nitrogen in organic and inorganic compounds. It involves three steps. Determination of relative proportions of different elements in both organic and inorganic compounds is known as quantitative analysis.

Given data:
Weight of the \[{\text{W = 2}}{\text{.8g}}\]
No. of moles of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4} = 20{\text{mmoles}}$

Complete step by step solution:
The Kjeldahl method was introduced by Johan Kjeldahl in 1883. This method generally involves three steps-digestion, distillation and titration. General equation to calculate the percentage of nitrogen is as given below:
\[\% \]of \[{\text{N = }}\dfrac{{{\text{1}}{\text{.4}} \times {\text{V}} \times {\text{N}}}}{{\text{W}}}\]
\[{\text{V}}\] is the volume of acid used for complete neutralization of evolved ammonia.
\[{\text{N}}\] is the normality of standard acid used.
\[{\text{W}}\] is the weight of a sample.
Here normality is not given.
Normality=Molarity*Basicity
Molarity is denoted as ${\text{M}}$.
Basicity is the number of protons it can donate. It is also called as n factor\[\left( {\text{n}} \right)\]
No. of protons that can be donated, \[{\text{n}} = 2\]
\[{\text{N}} = {\text{M}} \times {\text{n}}\]
Molarity is defined by number of moles solutes per litre of solution
No. of moles of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4} = 20{\text{mmoles}}$(Given)
$\therefore {\text{M}} = \dfrac{{20}}{{\text{V}}}$
Substituting the value of molarity in normality, we get
\[{\text{N}} = \dfrac{{20}}{{\text{V}}} \times 2 = \dfrac{{40}}{{\text{V}}}\]
Substituting the value of normality in the equation to find the \[\% \]of \[{\text{N}}\], we get,
\[\% \]of \[{\text{N}}\]\[ = \dfrac{{1.4 \times {\text{V}} \times \dfrac{{40}}{{\text{V}}}}}{{2.8{\text{g}}}}\]
Both volumes cancel each other.
\[\% \]of \[{\text{N}}\]$ = \dfrac{{1.4 \times 40}}{{2.8}} = 20\% $
Hence the option A is correct.

Additional information:
This method is helpful in determining nitrogen amount in soils, wastewater, fertilizers, etc. It can also be used in foods for determining the protein content. Oxidation of the organic compounds is the main purpose in this method.

Note: Using this method nitrogen cannot be converted to ammonium sulfate. This is because in some nitrogen containing compounds like azo and nitro groups or nitrogen containing rings, nitrogen cannot be quantitatively analyzed.